2012 AIPMT-MAINS Moving Charges and MagnetismMotion in Combined Electric and Magnetic Fields Medium Question 5 of 13 in Motion in Combined Electric and Magnetic Fields

A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an $\alpha$-particle to describe a circle of same radius in the same field?

A 4 MeV
B 2 MeV
C 1 MeV
D 0.5 MeV

The answer is C.

Solution

$R = \frac{mv}{qB} = \frac{\sqrt{2mE}}{qB}$
For equal radii: $\frac{\sqrt{2m_pE_p}}{q_pB} = \frac{\sqrt{2m_\alpha E_\alpha}}{q_\alpha B}$
$E_\alpha = \left(\frac{q_\alpha}{q_p}\right)^2\left(\frac{m_p}{m_\alpha}\right)E_p$, with $\frac{m_p}{m_\alpha} = \frac{1}{4}$ and $\frac{q_\alpha}{q_p} = 2$
$E_\alpha = (2)^2\times\frac{1}{4}\times E_p = E_p$
$E_\alpha = 1$ MeV

Concept Radius in terms of kinetic energy

Motion in Combined Electric and Magnetic Fields in past papers

13 questions from this chapter have appeared across 10 exam years.